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4 months ago

Binomial Theorem Class 11th

15. Find the expansion of (3x²– 2ax + 3a²)³using binomial theorem.

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Payal Gupta

Contributor-Level 10

15. ${[3 x^{2} - 2 a x + 3 a^{2}]}^{3}$

= ${[3 x^{2} + a (- 2 x + 3 a)]}^{3}$

We know that by binomial theorem,

${(a + b)}^{3}$ = $a^{3} + b^{3} + 3 a b (a + b)$

= $a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$

Then,

${[3 x^{2} + a (- 2 x + 3 a)]}^{3}$

= (3x²)³ + ${[a (- 2 x + 3 a)]}^{3}$ + $[3 (3 x^{2})^{2} a (- 2 x + 3 a)]$ + $[3 (3 x^{2}) {a (- 2 x + 3 a)}^{2}]$

= 27x⁶ + $[a^{3} {(- 2 x + 3 a)}^{3}]$ + $[3 (9 x^{4}) (- 2 a x + 3 a^{2})]$ + $[3 (3 x^{2}) {a^{2} {(3 a - 2 x)}^{2}}]$

= 27x⁶ + $[a^{3} {- 8 x^{3} + 27 a^{3} + 3 (4 x^{2}) (3 a) + 3 (- 2 x) (9 a^{2})}]$ + $[- 54 a x^{5} + 81 a^{2} x^{4}]$ + [ $(9 a^{2} x^{2})$ $(9 a^{2} + 4 x^{2} - 12 a x)$ ]

= 27x⁶ + [ $- 8 a^{3} x^{3} + 27 a^{6} + 36 a^{4} x^{2} - 54 a^{5} x$ ] + [ $- 54 a x^{5} + 81 a^{2} x^{4}$ ] + [ $(81 a^{4} x^{2} + 36 a^{2} x^{4} - 108 a^{3} x^{3}$ ]

= 27x⁶ $- 8 a^{3} x^{3} + 27 a^{6} + 36 a^{4} x^{2} - 54 a^{5} x$ $- 54 a x^{5} + 81 a^{2} x^{4}$ + $81 a^{4} x^{2} + 36 a^{2} x^{4} - 108 a^{3} x^{3}$

= 27x⁶– 54ax⁵ $+ 117 a^{2} x^{4}$ $- 116 a^{3} x^{3}$ + $117 a^{4} x^{2}$ $- 54 a^{5} x$ $+ 27 a^{6}$

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Binomial Theorem Class 11th

14. If a and b are distinct integers, prove that a – b is a factor of aⁿ– bⁿ, whenever n is a positive integer.

[Hint: write aⁿ= (a – b + b)ⁿ and expand]

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Payal Gupta

Contributor-Level 10

14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a - b + b)}^{n}$ = ${[(a - b) + b]}^{n}$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -}¹b + ⁿC₂(a – b)^{n –}²b² + ………… +ⁿC_n_-1 $(a - b) b^{n - 1}$ + ⁿC_nbⁿ

=>aⁿ= ${(a - b)}^{n}$ + ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + …………….…+ ⁿC_n_-1 $(a - b) b^{n - 1}$ + $b^{n}$ [Since, ⁿC₀ = 1 and ⁿC_n = 1]

=> $a^{n} - b^{n}$ = ${(a - b)}^{n}$ +ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + ……………… + ⁿC_n_-1 $(a - b) b^{n - 1}$

=> $a^{n} - b^{n}$ = $(a - b)$ [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +……….…… + ⁿC_n_-1 $b^{n - 1}$ ]

=> $a^{n} - b^{n}$ = $(a - b)$ k where k = [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +……….…… + ⁿC_n_-1 $b^{n - 1}$ ] is a natural number.

Therefore (a – b) is a factor o

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14. For (a – b) to be a factor of aⁿ – b ⁿwe need to show (aⁿ – bⁿ) = (a – b)k as k is a natural number.

We have, for positive n

aⁿ = ${(a - b + b)}^{n}$ = ${[(a - b) + b]}^{n}$

=>aⁿ = ⁿC₀(a – b)ⁿ + ⁿC₁(a – b)^{n -}¹b + ⁿC₂(a – b)^{n –}²b² + ………… +ⁿC_n_-1 $(a - b) b^{n - 1}$ + ⁿC_nbⁿ

=>aⁿ= ${(a - b)}^{n}$ + ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + …………….…+ ⁿC_n_-1 $(a - b) b^{n - 1}$ + $b^{n}$ [Since, ⁿC₀ = 1 and ⁿC_n = 1]

=> $a^{n} - b^{n}$ = ${(a - b)}^{n}$ +ⁿC₁ ${(a - b)}^{n - 1} b$ + ⁿC₂ ${(a - b)}^{n - 2} b^{2}$ + ……………… + ⁿC_n_-1 $(a - b) b^{n - 1}$

=> $a^{n} - b^{n}$ = $(a - b)$ [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +……….…… + ⁿC_n_-1 $b^{n - 1}$ ]

=> $a^{n} - b^{n}$ = $(a - b)$ k where k = [ ${(a - b)}^{n - 1}$ + ⁿC₁ ${(a - b)}^{n - 2} b$ + ⁿC₂ ${(a - b)}^{n - 3} b^{2}$ +……….…… + ⁿC_n_-1 $b^{n - 1}$ ] is a natural number.

Therefore (a – b) is a factor of aⁿ – bⁿwhere n is positive integer.

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Binomial Theorem Class 11th

13. Find a if the coefficients of x²and x³in the expansion of (3 + ax)⁹are equal.

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Payal Gupta

Contributor-Level 10

13. The general term of the expansion ${(3 + a x)}^{9}$ is

T_r₊₁ = ⁹C_r $3^{(9 - r)}$ ${(a x)}^{r}$

= ⁹C_r $3^{(9 - r)} a^{r} x^{r}$

At r = 2,

T₂₊₁ = ⁹C₂ $3^{(9 - 2)} a^{2} x^{2}$

= $\frac{9!}{2! (9 - 2)!}$ 3⁷a²x²

= $9′$ $8′$ $7!$ / $2′$ $1′$ $7!$ 3⁷a²x²

= 36 *3⁷a²x²

At r = 3,

T₃₊₁ = ⁹C₃ $3^{(9 - 3)} a^{3} x^{3}$

= $\frac{9!}{3! (9 - 3)!}$ 3⁶a³x³

= 9'8'7'6!/3'2'1'6! 3⁶a³x³

= 84 *3⁶a³x³

Given that,

Co-efficient of $x^{2}$ = co-efficient of $x^{3}$

=> 36 * $3^{7} a^{2}$ = 84 * $3^{6} a^{3}$

=> $\frac{a^{3}}{a^{2}}$ = 36' 3⁷^/84'3⁶

=> $a$ = $3′3/7$

= $\frac{9}{7}$

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Binomial Theorem Class 11th

12. Find a, b and n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively.

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Payal Gupta

Contributor-Level 10

1.The general term of the expansion (a + b)ⁿ is given by

T_r₊₁ = ⁿC_raⁿ^–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁aⁿ^-1b = $\frac{n!}{1! (n - 1)!}$ aⁿ^-1 b = $\frac{n * (n - 1)!}{(n - 1)!}$ aⁿ^-1b = naⁿ^-1b

T₃ = ⁿC₂aⁿ^-2b² = $\frac{n!}{2! (n - 2) [!}$ aⁿ^-2b² = $\frac{n * (n - 1) * (n - 2)!}{2 * 1 * (n - 2)!}$ aⁿ^-2b² = $\frac{n (n - 1)}{2}$ aⁿ^-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>naⁿ^–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b² = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

$\frac{n a^{n - 1} b}{a^{n}}$ = $\frac{7290}{729}$

=> $\frac{n b}{a}$ = 10

Similarly dividing equation (3) by (2) we get,

$\frac{n (n - 1)}{2}$ aⁿ^–2b² ÷ naⁿ^–1b = $\frac{30375}{7290}$

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b²* $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$

=> $n (n - 1) a^{n - 2} b^{2}$ * $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$ * 2

=> $\frac{(n - 1) b}{a}$ = $\frac{2 5′ * 2}{6}$

=> $\frac{n b}{a}$ $- \frac{b}{a}$ = $\frac{25}{3}$

=> 10 – $\frac{b}{a}$ = $\frac{25}{3}$ [since, $\frac{n b}{a}$ &

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1.The general term of the expansion (a + b)ⁿ is given by

T_r₊₁ = ⁿC_raⁿ^–rb^r

So, T₁ = ⁿC₀aⁿ = aⁿ

T₂ = ⁿC₁aⁿ^-1b = $\frac{n!}{1! (n - 1)!}$ aⁿ^-1 b = $\frac{n * (n - 1)!}{(n - 1)!}$ aⁿ^-1b = naⁿ^-1b

T₃ = ⁿC₂aⁿ^-2b² = $\frac{n!}{2! (n - 2) [!}$ aⁿ^-2b² = $\frac{n * (n - 1) * (n - 2)!}{2 * 1 * (n - 2)!}$ aⁿ^-2b² = $\frac{n (n - 1)}{2}$ aⁿ^-2b²

Given,

T₁ = 729

=>aⁿ = 729 ------------------ (1)

T₂ = 7290

=>naⁿ^–1b = 7290 ------------- (2)

T₃ = 30375

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b² = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

$\frac{n a^{n - 1} b}{a^{n}}$ = $\frac{7290}{729}$

=> $\frac{n b}{a}$ = 10

Similarly dividing equation (3) by (2) we get,

$\frac{n (n - 1)}{2}$ aⁿ^–2b² ÷ naⁿ^–1b = $\frac{30375}{7290}$

=> $\frac{n (n - 1)}{2}$ aⁿ^–2b²* $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$

=> $n (n - 1) a^{n - 2} b^{2}$ * $\frac{1}{n a^{n - 1} b}$ = $\frac{30375}{7290}$ * 2

=> $\frac{(n - 1) b}{a}$ = $\frac{2 5′ * 2}{6}$

=> $\frac{n b}{a}$ $- \frac{b}{a}$ = $\frac{25}{3}$

=> 10 – $\frac{b}{a}$ = $\frac{25}{3}$ [since, $\frac{n b}{a}$ = 10]

=> $\frac{b}{a}$ = 10 – $\frac{25}{3}$

= $\frac{30 - 25}{3}$

= $\frac{5}{3}$ --------------- (5)

Putting equation (5) in (4) we get,

n* $\frac{5}{3}$ = 10

=>n = 10 * $\frac{3}{5}$

=>n = 6

So putting the value of n in equation (1) we get,

a6 = 729

=>a⁶= 36

=>a = 3

And putting a = 3 in equation (5) we get,

$\frac{b}{a}$ = $\frac{5}{3}$

=>b = $\frac{5}{3}$ *a

= $\frac{5}{3}$ * 3

= 5

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Binomial Theorem Class 11th

11. Find a positive value of m for which the coefficient of x²in the expansion (1 + x)^mis 6.

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Payal Gupta

Contributor-Level 10

11. The general term of the expansion ${(1 + x)}^{m}$ is given by,

T_r₊₁ = ^mC_r $1^{m - r} x^{r}$

= ^mC_rx^r

At r = 2,

T₂₊₁ = ^mC₂x²

Given that, co-efficient of x² = 6

=>^mC₂ = 6

=> $\frac{m!}{2! (m - 2)!}$ = 6

=>m² – m = 12

=>m² – m – 12 = 0

=>m² + 3m – 4m – 12 = 0

=>m (m + 3) – 4 (m+ 3) = 0

=> (m – 4) (m + 3) = 0

=>m = 4 and m = –3

Since, we need a positive value of m we have, m = 4

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Binomial Theorem Class 11th

10. Prove that the coefficient of xⁿin the expansion of (1 + x)²ⁿis twice the coefficientof xⁿin the expansion of (1 + x)^2n-1.

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Payal Gupta

Contributor-Level 10

10. General term of the expansion (1 + x)²ⁿ is

T_r₊₁ = ²ⁿC_r (1)^2n-r(x)^r

So, co-efficient of xⁿ (i.e. r = n) is ²ⁿC_n

Similarly general term of the expansion (1 + x)^2n–1 is

T_r₊₁ = ^2n-1C_r (1)^2n–1–rx^r

And co-efficient of xⁿ i.e. when r = n is ^2n-1C_n

Therefore,

$\frac{c o - e f f i c i e n t o f x^{n} i n {(1 + x)}^{2 n}}{c o - e f f i c i e n t o f x^{n} i n {(1 + x)}^{2 n - 1}}$

= $\frac{2 n!}{n! (2 n - n)!}$ ÷ $\frac{(2 n - 1)!}{n! (2 n - 1 - n)!}$

= $\frac{2 n!}{n! n!}$ * $\frac{n! (n - 1)!}{(2 n - 1)!}$

= $\frac{2 n}{n}$

= 2

Thus, co-efficient of $x^{n}$ in ${(1 + x)}^{2 n}$ = 2x co-efficient of $x^{n}$ in ${(1 + x)}^{2 n - 1}$

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Binomial Theorem Class 11th

9. The coefficients of the (r – 1)^th, r^thand (r + 1)^thterms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find n and r.

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Payal Gupta

Contributor-Level 10

9. The general term of the expansion (x +1)ⁿ is

T_r₊₁ = ⁿC_rxⁿ^–r1^r

i.e. co-efficient of ${(r + 1)}^{t h}$ term = ⁿC_r

So, co-efficient of ${(r - 1)}^{t h}$ term =ⁿC_{(r–1) – 1} = ⁿC_r_{– 2}

Similarly, co-efficient of rth term = ⁿC_r_{– 1}

Given that, ⁿC_r_{– 2}:ⁿC_r_{– 1}: ⁿC_r = 1 : 3 : 5

We have,

= $\frac{1}{3}$

=> $\frac{n!}{(r - 2)! (n - r + 2)!}$ * $\frac{(r - 1)! (n - r + 1)!}{n!}$ = $\frac{1}{3}$

=> $\frac{(r - 1) (r - 2)! (n - r + 1)!}{(r - 2)! (n - r + 2) (n - r + 1)!}$ = $\frac{1}{3}$

=> $\frac{(r - 1)}{(n - r + 2)}$ = $\frac{1}{3}$

=> 3r – 3 = n – r + 2

=> 3r + r = n + 2 + 3

=> 4r = n + 5 -------------- (1)

And,

= $\frac{3}{5}$

=> $\frac{n!}{(r - 1)! (n - r + 1)!}$ * $\frac{r! (n - r)!}{n!}$ = $\frac{3}{5}$

=> $\frac{r (r - 1)! (n - r)!}{(r - 1)! (n - r + 1) (n - r)!}$ = $\frac{3}{5}$

=> $\frac{r}{n - r + 1}$ = $\frac{3}{5}$

=> 5r = 3n – 3r + 3

=> 5r + 3r = 3n + 3

=> 8r = 3n + 3 ----------------------- (2)

Multiplying equation (

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9. The general term of the expansion (x +1)ⁿ is

T_r₊₁ = ⁿC_rxⁿ^–r1^r

i.e. co-efficient of ${(r + 1)}^{t h}$ term = ⁿC_r

So, co-efficient of ${(r - 1)}^{t h}$ term =ⁿC_{(r–1) – 1} = ⁿC_r_{– 2}

Similarly, co-efficient of rth term = ⁿC_r_{– 1}

Given that, ⁿC_r_{– 2}:ⁿC_r_{– 1}: ⁿC_r = 1 : 3 : 5

We have,

= $\frac{1}{3}$

=> $\frac{n!}{(r - 2)! (n - r + 2)!}$ * $\frac{(r - 1)! (n - r + 1)!}{n!}$ = $\frac{1}{3}$

=> $\frac{(r - 1) (r - 2)! (n - r + 1)!}{(r - 2)! (n - r + 2) (n - r + 1)!}$ = $\frac{1}{3}$

=> $\frac{(r - 1)}{(n - r + 2)}$ = $\frac{1}{3}$

=> 3r – 3 = n – r + 2

=> 3r + r = n + 2 + 3

=> 4r = n + 5 -------------- (1)

And,

= $\frac{3}{5}$

=> $\frac{n!}{(r - 1)! (n - r + 1)!}$ * $\frac{r! (n - r)!}{n!}$ = $\frac{3}{5}$

=> $\frac{r (r - 1)! (n - r)!}{(r - 1)! (n - r + 1) (n - r)!}$ = $\frac{3}{5}$

=> $\frac{r}{n - r + 1}$ = $\frac{3}{5}$

=> 5r = 3n – 3r + 3

=> 5r + 3r = 3n + 3

=> 8r = 3n + 3 ----------------------- (2)

Multiplying equation (1) by 2 and subtracting from equation (2) we get,

8r – 8r = 3n + 3 – (2n + 10)

=> 0 = 3n + 3 – 2n – 10

=> 0 = n – 7

=>n = 7

Putting n = 7 in equation (1) we get,

=> 4r = 7 + 5

=>r = $\frac{12}{4}$

=>r = 3

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Binomial Theorem Class 11th

8. In the expansion of (1 + a)^m⁺ⁿ,prove that coefficients of aⁿ and aⁿ are equal.

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Payal Gupta

Contributor-Level 10

8. The general term of the expansion (1 + a)^m⁺ⁿ is

T_r₊₁ = ^m⁺ⁿC_ra^r [since, 1^m^+n-r = 1]

At r = m we have,

T_m₊₁ = ^m⁺ⁿC_ma^m

= $\frac{(m + n)!}{m! (m + n - m)!}$ (a)^m

= $\frac{(m + n)!}{m! n!}$ a^m - (1)

Similarly at r = n we have,

T_n₊₁ = ^m⁺ⁿC_naⁿ

= $\frac{(m + n)!}{n! (m + n - n)!}$ (a)ⁿ

= $\frac{(m + n)!}{n! m!}$ aⁿ - (2)

Hence from (1) & (2),

Co-efficient of am = Co-efficient of aⁿ = $\frac{(m + n)!}{m! n!}$

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Binomial Theorem Class 11th

7. Find the 4thterm in the expansion of (x – 2y)¹².

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Payal Gupta

Contributor-Level 10

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Binomial Theorem Class 11th

6. ( $x^{2} - y x)^{12}$ , x≠ 0.

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Payal Gupta

Contributor-Level 10

6. Let (r + 1)th be the general term of ( $x^{2} - y x)^{12}$

So, T_r_-1 = ¹²C_r (x²)^12–r (–yx)^r

= (–1)^r¹²C_rx^24–2ry^rx^r

= (–1)^r¹²C_r $x^{24 - 2 r + r} y^{r}$

= (-1)^r¹²C_r $x^{24 - r} y^{r}$

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